Download Mathematics 220 Homework for Week 7 Due March 6 If

Mathematics 220
Homework for Week 7
Due March 6
If you are using the 2nd edition, be careful — question numbers may not agree.
5.4 Disprove the statement: Let n ∈ N. If
Solution: Let n = 2, then
n = 2 is a counterexample.
(n + 1)(n + 2)
n(n + 1)
is odd, then
is odd.
2
2
n(n + 1)
(n + 1)(n + 2)
= 3 is odd but
= 6 is even. So
2
2
5.16 Prove that the product of an irrational number and a nonzero rational number is
rational.
Solution: Assume, to the contrary, that there exists a nonzero rational number
r and an irrational number x, so that rx is rational. Since r and rx are rational,
c
a
there are integers a, b, c, d with b and d nonzero, so that r = and rx = . Then
b
d
a
c
x=
b
d
If we multiply both sides by ab we have
a
b
c
b
x=
a
b
a
d
⇒
x=
bc
ad
But bc and ad are products of integers and are integers; so x is rational, which
contradicts the assumption.
5.24 Prove that there exists no positive integer x such that 2x < x2 < 3x.
Solution: Assume, to the contrary, that x is a positive integer with 2x < x2 < 3x.
Since x is positive. 1/x is positive. Hence
2x(1/x) < x2 (1/x) < 3x(1/x)
⇒
2 < x < 3.
But there is no integer between 2 and 3 and this contradicts the assumption that
x is an integer.
5.32 Prove that there exist no positive integers m and n for which m2 + m + 1 = n2 .
Solution: Assume, to the contrary, that m and n are positive integers with m2 +
m + 1 = n2 . Since m is positive
m2 < m2 + m + 1 = n2 < m2 + 2m + 1 = (m + 1)2 .
Mathematics 220
Homework for Week 7
Due March 6
Because m, n and m + 1 are positive, from the above inequality we conclude that
m < n < m + 1.
But there is no integer which is strictly between m and m + 1. This contradicts the
assumption that n is an integer and proves the statement.
5.36 Let a, b ∈ R. Prove that if ab 6= 0, then a 6= 0 by using as many of the three proof
techniques as possible.
Solution: Direct Proof: Suppose a, b ∈ R are given and ab 6= 0. We cannot have
ab
b = 0 otherwise ab = 0 which is impossible, so b 6= 0. Because b 6= 0 then a =
is
b
nonzero.
Proof by contrapositive: We prove the contrapositive: if a = 0 then ab = 0.
The product of every real number with zero is zero. So if a = 0 then ab = 0.
Proof by contradiction: Assume, to the contrary, that ab 6= 0 but a = 0 but the
product of every real number with zero is zero; this is a contradiction and proves
the statement.
5.40 Show that there exist a rational number a and an irrational number b such that ab is
rational.
Solution: Let a = 1 and b =
√
2; then ab = 1 is rational.
6.2 Prove that if A is any well-ordered set of real numbers and B is a nonempty subset of
A, then B is also well-ordered.
Solution: Assume A ⊂ R is well-ordered and B is a nonempty subset. Given a
nonempty subset C of B, C is also a nonempty subset of A. Since A is well-ordered,
C has a least element. So we have showed that every nonempty subset of B has a
least element. Hence B is well-ordered.
6.8 Find a formula for 1 + 4 + 7 + · · · + (3n − 2) for positive integer n, and then verify your
formula by mathematical induction.
Solution: We use induction to prove
1 + 4 + 7 + · · · + (3n − 2) =
Page 2
n(3n − 1)
.
2
Mathematics 220
Homework for Week 7
Due March 6
When n = 1, obviously both hand sides of the above identity are equal to 1; so the
statement holds for n = 1. Assume the identity holds for k a positive integer:
1 + 4 + 7 + · · · + (3k − 2) =
n(3k − 1)
,
2
we use that to show the statement holds for k + 1:
1 + 4 + 7 + · · · + (3k − 2) + [3(k + 1) − 2] = [1 + 4 + 7 + · · · + (3k − 2)] + (3k + 1)
n(3k − 1)
=
+ (3k + 1)
2
3k 2 − k + 2(3k + 1)
=
2
3k 2 + 5k + 2
=
2
(k + 1)(3k + 2)
=
2
(k + 1)[(3(k + 1) − 1]
=
.
2
This shows that the identity holds for k + 1. By the principle of mathematical
induction, this proves the statement for every positive integer n.
6.12 Consider the open sentence P (n) : 9 + 13 + · · · + (4n + 5) =
n ∈ N.
4n2 + 14n + 1
, where
2
(a) Verify the implication P (k) ⇒ P (k + 1) for an arbitrary positive integer k.
(b) Is ∀n ∈ N, P (n) true?
Solution:
(a) Assume P (k) holds, namely:
9 + 13 + · · · + (4k + 5) =
4k 2 + 14k + 1
,
2
we want to prove P (k + 1):
9 + 13 + · · · + (4k + 5) + [4(k + 1) + 5] =
4(k + 1)2 + 14(k + 1) + 1
.
2
We start from the left hand side and use P (k):
4k 2 + 14k + 1
+ (4k + 9)
2
4k 2 + 14k + 1 + 8k + 18
=
2
4k 2 + 22k + 19
=
2
9 + 13 + · · · + (4k + 5) + [4(k + 1) + 5] =
Page 3
Mathematics 220
Homework for Week 7
Due March 6
On the other hand for the right hand side of P (k + 1) we have:
4(k + 1)2 + 14(k + 1) + 1
4(k 2 + 2k + 1) + 14k + 14 + 1
4k 2 + 22k + 19
=
=
.
2
2
2
This shows the two sides of the identity in P (k + 1) are equal and therefore we
have proved P (k + 1).
(b) No! Let n = 1, the left hand side of the equality in P (1) has only one term and
is equal to 9. But for the right hand side, we have
4 × 1 + 14 × 1 + 1
19
=
2
2
which is not equal to 9.
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